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%%文档的题目、作者与日期
\author{王立庆（2020级数学与应用数学1班） }
\title{随机分析入门习题解答 -- 鞅的概念}
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%\date{2021 年 9 月 14 日}
%\date{March 9, 2021}

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\begin{enumerate}\itemsep1em

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\item  %第1题：多项选择
Find the correct statements about discrete-time martingales. 
\begin{enumerate}
\item[a.]  A discrete-time martingale is a discrete-time stochastic process $X=\{X_n,\, n\ge 0\}$. 
\item[b.]  A discrete-time martingale also consists of a filtration of $\sigma$-fields $\{\mathcal{F}_n,\, n\ge 0\}$. 
\item[c.]  For a martingale, the stochastic process is adapted to the filtration, i.e., $\sigma(X_n)\subseteq\mathcal{F}_n$. 
\item[d.]  For a martingale $(X,\{\mathcal{F}_n\} )$, for $s>t$, the best prediction of $X_t$ given $\mathcal{F}_s$ is $X_s$. 
\end{enumerate} 

\vspace{0.2cm}

{\color{red}解答：abc. 离散时间的鞅是一个离散时间的随机过程以及一个事件域的序列，符合一定的条件。选项 d 中的 $s>t$ 改成 $s<t$ 就对了。设 $s<t$, 在已知当前的 $s$ 时刻的信息，对于鞅过程 $X$ 在将来的 $t$ 时刻的值 $X_t$, 最好的预测是当前的值 $X_s$. 写成公式就是 $\mathbb{E}(X_t | \mathcal{F}_s) = X_s$. 

}

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\item  %第2题：多项选择
Let $B=\{B_t,\, t\ge 0\}$ be a standard Brownian motion. Let $\mathcal{F}_t=\sigma(B_s,\, s\le t)$ be the natural filtration. Find the correct statements. 
\begin{enumerate}
\item[a.]  The Brownian motion $B$ is a martingale with respect to the filtration $\{\mathcal{F}_t\}$. 
\item[b.]  The stochastic process $\{B_t^2-t,\, t\ge 0\}$ is a martingale with respect to the filtration $\{\mathcal{F}_t\}$. 
\item[c.]  Let $s<t$. Since $B_t-B_s$ and $B_s$ are independent, we have $\mathbb{E}(B_t-B_s) | B_s) = \mathbb{E}(B_t-B_s)=0$. 
\item[d.]  The $\sigma$-fields $\{\mathcal{F}_t,\, t\ge 0\}$ form an increasing stream of information about the process $B$.  
\end{enumerate} 

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{\color{red}解答：abcd. 从布朗运动可以自然地得到一个连续时间的信息不断增加的事件域的序列。布朗运动相对于这个事件域序列而言是一个连续时间的鞅过程。选项 c 是在说，因为布朗运动是一个独立增量的随机过程，而且每段时间上的增量是一个均值为零的正态分布，所以布朗运动符合鞅过程的定义中的关键条件，即 $\mathbb{E}(B_t | \mathcal{F}_s) = B_s$ 对任意 $s<t$ 都成立。

}

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\item  %第3题：多项选择
Let $X=\{X_n,\,n\ge 0\}$ be a discrete-time martingale with respect to a filtration $\{\mathcal{F}_n,\, n\ge 0\}$. 
Let $C=\{C_n,\, n\ge 1\}$ be a discrete-time stochastic process. 
The following four steps define the martingale transform. 
Find the step that is incorrect. 
\begin{enumerate}
\item[a.]  Form the martingale difference sequence $Y_{n+1}=X_{n+1}-X_n,\, n\ge 0$. 
\item[b.]  Assume that the information carried by $C_n$ is contained in $\mathcal{F}_{n}$ for all $n\ge 1$. 
\item[c.]  Define a stochastic process $Z$ by $Z_0=0$ and $Z_n=C_1Y_1+\cdots+C_nY_n$ for all $n\ge 1$.  
\item[d.]  Assume $\mathbb{E}(C_n^2)<\infty$ and $\mathbb{E}(Y_n^2)<\infty$ for all $n$. Prove that $Z$ is a martingale. 
\end{enumerate} 

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{\color{red}解答：b. 为使得鞅变换的结果仍是鞅过程，随机过程 $C$ 需要满足关于给定的事件域序列是可预测的条件。因此选项 b 要修改成：随机变量 $C_n$ 生成的 $\sigma$-域是事件域 $\mathcal{F}_{n-1}$ 的子集，而不仅是  $\mathcal{F}_{n}$ 的子集。 

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\item  %第4题：多项选择
The conditional expectation has several useful rules. One of them is this. If $X$ and $\mathcal{F}$ are independent, then $\mathbb{E}(X | \mathcal{F}) = \mathbb{E}(X)$. The following four statements try to explain this. Find the statement that is incorrect. 
\begin{enumerate}
\item[a.]  By assumption, the random variable $X$ and the $\sigma$-field $\mathcal{F}$ are independent. 
\item[b.]  For all events $A\in\mathcal{F}$, the random variables $X$ and $I_A$ are independent. 
\item[c.]  Denote $m=\mathbb{E}(X)$. This is a constant. Since $X$ and $I_A$ are independent, we have $$\mathbb{E}(XI_A)=\mathbb{E}(X)\mathbb{E}(I_A)=\mathbb{E}[mI_A].$$
\item[d.]  The constant $m$ contains more information than any $\sigma$-field. 

\end{enumerate} 

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{\color{red}解答：d. 常数生成的 $\sigma$-域是平凡的，即 $\sigma(m)=\{\varnothing, \Omega\}$. 它包含的信息比其它 $\sigma$-域都要来得少。因为这一点，以及选项 c 中的等式对任意 $A\in\mathcal{F}$ 都成立，所以“常数随机变量” $m$ 就是条件期望 $\mathbb{E}(X | \mathcal{F})$. 

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\item  %第5题：多项选择
Let $B=\{B_t,\, t\ge 0\}$ be the standard Brownian motion. Let $\mathcal{F}_t=\sigma(B_s,\, s\le t)$ be the natural filtration. 
Find the stochastic processes that are adapted to the filtration $\{\mathcal{F}_t,\, t\ge 0\}$.
\begin{enumerate}
\item[a.]  $\{X_t=B_t^2-t,\,\, t\ge 0\}$. 
\item[b.]  $\{X_t = B_{t+2},\,\, t\ge 0\}$.
\item[c.]  $\{X_t = \max\{B_s\mid 0\le s\le t\}, \,\, t\ge 0\}$. 
\item[d.]  $\{X_t = B_T - B_t,\,\, t\ge 0\}$ for some fixed $T>0$. 
\end{enumerate} 

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{\color{red}解答：ac. 在选项 b 和 d 中，随机过程 $X_t$ 在 $t$ 时刻的值与布朗运动 $\{B_t\}$ 在 $t$ 时刻之后的信息有关。因此这两个随机过程不是适应于布朗运动 $\{B_t\}$ 生成的事件域序列的。

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\item  %第6题：多项选择
Let $B=\{B_t,\, t\ge 0\}$ be the standard Brownian motion. Let $\mathcal{F}_t=\sigma(B_s,\, s\le t)$ be the natural filtration. 
The stochastic process $\{X_t = B_t^2,\,\, t\ge 0\}$ generates another filtration $\{ \mathcal{F}_t' = \sigma(B_s^2, s\le t), \,\, t\ge 0\}$. What can we say about these two filtrations?
\begin{enumerate}
\item[a.]  For every $t$, the $\sigma$-field $\mathcal{F}'_t$ is a subset of the $\sigma$-field $\mathcal{F}_t$. 
\item[b.]  From the process $\{X_t\}$ we can only recover the information about $|B_t|$.
\item[c.]  The process $\{X_t\}$ is adapted to the natural filtration of the Brownian motion.
\item[d.]  In finance there are people who know more information, and thus they have their own filtration which can be bigger than the natural filtration representing the market fluctuation. 
\end{enumerate} 

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{\color{red}解答：abcd. 选项 a 是对的，因为随机变量 $B_t^2$ 生成的事件域是 $B_t$ 生成的事件域的一个子集。选项 b 是对的，因为从 $B_t^2$ 我们无法知道 $B_t$ 的正负号。选项 c 是对的，因为选项 a 是对的。选项 d 是对的，因为在金融市场中，知道更多信息的人，在同一时刻，比别人拥有更大的事件域。

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\item  %第7题：多项选择
What can we say about the conditional expectation $\mathbb{E}(X_t\mid\mathcal{F}_s)$ for $0\le s<t$?
\begin{enumerate}
\item[a.]  If $\mathcal{F}_s$ and $X$ are dependent, we can expect to reduce the uncertainty of $X_t$. 
\item[b.]  If $\mathcal{F}_s$ and $X_t$ are dependent, the conditional expectation $\mathbb{E}(X_t\mid\mathcal{F}_s)$ is $X_s$. 
\item[c.]  If we know certain events happened before or at time $s$, we can predict $X_t$ better with the information in $\mathcal{F}_s$ than without it. 
\item[d.]  The conditional expectation $\mathbb{E}(X_t\mid\mathcal{F}_s)$ is the best prediction of $X_t$ given the information $\mathcal{F}_s$. 

\end{enumerate} 

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{\color{red}解答：acd. 选项 b 是不对的，因为随机过程 $X$ 相对于递增的事件域序列 $\{\mathcal{F}_s\}$ 没有说是鞅。条件期望 
$\mathbb{E}(X_t\mid\mathcal{F}_s)$ 是一个数学模型，描述了如何用已知的信息对将来发生的结果作出预测。

}

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\item  %第8题：多项选择
Find the correct statements about continuous-time martingales. 
\begin{enumerate}
\item[a.]  A continuous-time martingale is a continuous-time stochastic process $X=\{X_t,\, t\ge 0\}$. 
\item[b.]  A continuous-time martingale also consists of a filtration of $\sigma$-fields $\{\mathcal{F}_t,\, t\ge 0\}$. 
\item[c.]  For a martingale, the stochastic process is adapted to the filtration, i.e., $\sigma(X_t)\subseteq\mathcal{F}_t$. 
\item[d.]  For a martingale $\{X_t,\mathcal{F}_t\}$, for $s>t$, the best prediction of $X_t$ given $\mathcal{F}_s$ is $X_s$. 
\end{enumerate} 

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{\color{red}解答：abcd. 这四条组成了连续时间鞅的定义。我们说一个随机过程是一个鞅过程，总是相对于某个递增的事件域序列而言的。

}

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\item  %第9题：多项选择
Let $\{Z_n\}$ be a sequence of independent random variables with expectations all zero. Consider the partial sums $R_n=Z_0+Z_1+\cdots+Z_n, \,\, n\ge 0$. Consider the natural filtration $\mathcal{F}_n=\sigma(R_0, R_1,\cdots, R_n)$. 
Find the correct statements.
\begin{enumerate}
\item[a.]  $\{\mathcal{F}_n\}$ is an increasing sequence of $\sigma$-fields. 
\item[b.]  The $\sigma$-field $\mathcal{F}_n$ also equals to $\sigma(Z_0, Z_1,\cdots, Z_n)$. 
\item[c.]  Since the $\{Z_n\}$ are independent, the $\sigma$-field $\mathcal{F}_n$ and $Z_{n+1}$ are independent. 
\item[d.]  Since $\mathbb{E}(R_{n+1}\mid \mathcal{F}_n)=R_n$, assuming $\mathbb{E}|Z_n|<\infty$, the process $\{R_n\}$ is a martingale w.r.t. $\{\mathcal{F}_n\}$. 
\end{enumerate} 

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{\color{red}解答：abcd. 这是由相互独立的均值都为零的一列随机变量的部分和组成的随机过程。可以验证离散时间鞅过程的定义里的条件都是成立的。

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\item  %第10题：多项选择
From a random variable $Z$ and a filtration $\{\mathcal{F}_n\}$, we may construct a martingale. Find the correct statements. 
\begin{enumerate}
\item[a.]  The filtration $\mathcal{F}_0\subseteq \mathcal{F}_1\subseteq \cdots\subseteq \mathcal{F}_n\subseteq \cdots$ represents the fact that more and more information are gained about the random variable $Z$. 
\item[b.]  Define a stochastic process $X_n = \mathbb{E}(Z\mid \mathcal{F}_n)$ for each $n\ge 0$. 
\item[c.]  Assume $\mathbb{E}|Z|<\infty$. Then the sequence $\{X_0,X_1,\cdots,X_n,\cdots\}$ is a martingale w.r.t. $\{\mathcal{F}_n\}$.
\item[d.]  If $\sigma(Z)\subseteq \mathcal{F}_n$ for some $n$, then $X_m=Z$ for all $m\ge n$.
\end{enumerate} 

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{\color{red}解答：abcd. 设 $\mathcal{F}_0=\{\varnothing,\Omega\}$, 则 $X_0=\mathbb{E}(Z\mid \mathcal{F}_0) =\mathbb{E}(Z)$. 也就是说，在时刻 $t=0$, 我们只知道随机变量 $Z$ 的均值。设 $\mathcal{F}_1=\{\varnothing,\Omega, A,A^c\}$, 则 $X_1=\mathbb{E}(Z\mid \mathcal{F}_1)$ 在 $\mathbb{E}(Z)$ 的基础上又给出了 $\mathbb{E}(Z\mid A)$ 与 $\mathbb{E}(Z\mid A^c)$ 的信息。直到 $\sigma(Z)$ 全都落在某个 $\mathcal{F}_n$ 之内，这时候 $X_n$ 就完全包含 $Z$ 的所有情况了。

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\end{enumerate}

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